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So the big theory is every hilbert space operator is a linear combination of projections, and for matrices, Any Hermitian Matrix is the Linear Combination of Four Projections.

But the proof in the paper does not give a explicit algorithm to compute those four projections, thus I am curious how we should look for those four projections.

Thanks!

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Do you really think $[0,0;1,0]$ is Hermitian? –  Did Jul 7 '12 at 14:12
    
@did I am asking about general matrices, which are linear combinations of hermitian matrices. –  Hui Yu Jul 7 '12 at 14:31
    
@did Since any matrix is the sum of its real part and imaginary part, both of which are Hermitian. Such a result for Hermitian matrices would automatically apply to general matrices. –  Hui Yu Jul 7 '12 at 14:55
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@did By real part he means $\frac{A+A^*}2$. –  azarel Jul 7 '12 at 16:35
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A self-adjoint $n$-by-$n$ matrix is a linear combination of at most $n$ commuting projections, which is probably more useful. Normal matrices can be characterized as linear combinations of commuting projections. –  Jonas Meyer Jul 8 '12 at 3:39
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V I Rabanovich, On the decomposition of a diagonal operator into a linear combination of idempotents or projections, Ukrainian Math. J. 57 (2005), no. 3, 466–473, MR2188435 (2006g:47001), writes, "We prove (Theorem 2) that a diagonal self-adjoint operator is a linear combination of four orthoprojectors." I haven't tried to work through the details, but I think there may be enough there to answer your question.

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